Question

If one zero of the polynomial (a²+9)x²+13x+6a is reciprocal of other, find a. ​

Answer 1

Answer:

value of a is 3.

Step-by-step explanation:

Given polynomial:

• (a²+9)x² + 13x + 6a

Comparing the given polynomial with ax² + bx + c, we get,

• a = a²+9
• b = 13
• c = 6a

Let one zero be m.

other zero = 1/m

We know that,

product of zeroes = c/a

=> m × 1/m = 6a/(a²+9)

=> 1 = 6a/(a²+9)

=> a²+9 = 6a

=> a² - 6a + 9 = 0

By splitting middle term,

=> a² - 3a - 3a + 9 = 0

=> a ( a - 3 ) - 3 ( a - 3 ) = 0

=> ( a - 3 ) ( a - 3 ) = 0

=> a = 3

Therefore, value of a is 3.

Answer 2

Answer:

Step-by-step explanation:

Every quadratic equation has two roots ;either real or complex.

for a given quadratic equation ax²+bx+c =0 if the roots are e and f then

e+f= (-b)/a      e×f= c/a

so here in the given question one root is the reciprocal of the other i.e if a roots is a the the other root is 1/a.

so the product of the roots is 1.

i.e c/a=1

   (6a/(a²+9))=1   or 6a=a²+9  or a²-6a+9=0

i.e      (a-3)²=0

     so   a-3=0  or a=3